AP EAMCET · Maths · Circle
A circle is such that \((x-2) \cos \theta+(y-2) \sin \theta=1\) touches it for all values of \(\theta\). Then, the circle is
- A \(x^2+y^2-4 x-4 y+7=0\)
- B \(x^2+y^2+4 x+4 y+7=0\)
- C \(x^2+y^2-4 x-4 y-7=0\)
- D \(x^2+y^2+4 x+4 y-7=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-4 x-4 y+7=0\)
Step-by-step Solution
Detailed explanation
Since, the line \((x-2) \cos \theta+(y-2) \sin \theta=1\) touches a circle. So it is a tangent equation to a circle. Equation of tangent to a circle at \(\left(x_1, y_1\right)\) is \((x-h) x_1+(y-k) y_1\) \(=a^2\) to a circle \((x-h)^2+(y-k)^2=a^2\) Then, after comparing…
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